# Prove that a limit does not exist

• Anro
In summary, using the sequential criterion for nonexistence, we can prove that limx→2 (1/(x – 2)) does not exist by finding two sequences that converge to 2, but have different limits of the function at that point. This shows that the limit of the function does not exist at x = 2.

## Homework Statement

Prove that limx→2 (1/(x – 2)) does not exist.

2. The attempt at a solution

My approach to this problem is to use the sequential criterion for nonexistence, that is, if {x_n} and {y_n} both converge to a (2 in this case), but limx→a f(x_n) is not equal to limx→a f(y_n), then limx→a f(x) does not exist (a being 2 in this case).

I was able to find two sequences, {x_n} and {y_n}, both converging to 2 (I chose {x_n} = 1.5 + (1/n) and {y_n} = 2.5 – (1/n) as converging sequences). But when I try to plug these values in f(x_n) and f(y_n), I get (2 – n) or an integral multiple of it in the denominator which doesn’t help me when taking the limit for either f(x_n) or f(y_n) when n→2.

I also tried some other combinations but I seem to run into the same problem, division by zero when taking the limit for either f(x_n) or f(y_n), can you please tell me if I am missing something or if I am doing something wrong here? Thanks in advance for all your help.

Anro said:

## Homework Statement

Prove that limx→2 (1/(x – 2)) does not exist.

2. The attempt at a solution

My approach to this problem is to use the sequential criterion for nonexistence, that is, if {x_n} and {y_n} both converge to a (2 in this case), but limx→a f(x_n) is not equal to limx→a f(y_n), then limx→a f(x) does not exist (a being 2 in this case).

I was able to find two sequences, {x_n} and {y_n}, both converging to 2 (I chose {x_n} = 1.5 + (1/n) and {y_n} = 2.5 – (1/n) as converging sequences). But when I try to plug these values in f(x_n) and f(y_n), I get (2 – n) or an integral multiple of it in the denominator which doesn’t help me when taking the limit for either f(x_n) or f(y_n) when n→2.

I also tried some other combinations but I seem to run into the same problem, division by zero when taking the limit for either f(x_n) or f(y_n), can you please tell me if I am missing something or if I am doing something wrong here? Thanks in advance for all your help.
You don't want n → 2. You need {xn} → 2 as n → ∞, and {yn} → 2 as n → ∞, while f(x_n) ≠ f(y_n) as n → ∞.

What you have done (n->infinity, $x_n$-> 2) is perfectly valid (I think Sammy S misunderstood what you were saying) but, since you get $x_n-2$ in the denominator, the denominator is going to 0 while the numerator is always 1. Of course, that does not exist- it goes to infinity.

(Note: saying $lim_{x\to 2} f(x)= \infty$ is just another way of say the limit does not exist. "Infinity" is NOT a number so is not a limit.)

Or, try $x_n= 2+ 1/n$, $y_n= 2- 1/n$. As n goes to infinity, $x_n$ goes to 2 "from above" and the limit is +infinity while $y_n$ goes to 2 "from below" and the limit is -infinity.

it is possible to use the definition of the limit directly, here.

suppose that f DID have a limit at 2, call it L.

could this limit be 0? no, because if we choose, say ε = 1,

then anytime |x-2| < 1, |f(x)| = 1/|x-2| > 1, so we can't find a δ.

so if L exists, L ≠ 0. therefore |L| > 0, and we can use that as our ε.

but if 0 < |x-2| < 1/(2|L|),

then |1/(x-2) - L| ≥ 1/|x-2| - |L|...can you see how to finish?

no matter how small a δ we choose, there will be some x for which |x-2| < 1/(2|L|),

which means...?

Thanks for the answers everyone; here is what I have done so far:

Choose {x_n} = 2 + (1/n). Then limn→∞ (x_n) = 2.
Hence 1/((x_n) – 2) = n = f(x_n), and limn→∞ f(x_n) = limn→∞ n = ∞

Now choose {y_n} = 2 - (1/n). Then limn→∞ (y_n) = 2.
Hence 1/((y_n) – 2) = (-n) = f(y_n), and limn→∞ f(y_n) = limn→∞ (-n) = -∞

Since limn→∞ f(x_n) does not equal limn→∞ f(y_n), then limn→∞ f(x) does not exist.

Thus limx→2 (1/(x – 2)) does not exist.